All posts for the month July, 2015

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by and then employing the substitution (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

So far, I have shown that

where and (and is a certain angle that is now irrelevant at this point in the calculation).

There are actually a couple of ways for computing this last integral. Today, I’ll lay the foundation for the “magic substitution”

With this substitution, the above integral will become a rational function, which can then be found using standard techniques.

First, we use some trig identities to rewrite in terms of :

Next, I’ll replace by :

.

Second, for the sake of completeness (even though it isn’t necessary for this particular integral), I’ll rewrite in terms of :

Next, I’ll replace by :

.

Third, again for the sake of completeness,

.

Finally, I need to worry about what happens to the :

These four substitutions can be used to convert trigonometric integrals into some other integral. Usually, the new integrand is pretty messy, and so these substitutions should only be used sparingly, as a last resort.

I’ll continue this different method of evaluating this integral in tomorrow’s post.

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by and then employing the substitution (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

So far, I have shown that

.

where and (and is a certain angle that is now irrelevant at this point in the calculation).

I now write as a new sum by again dividing the region of integration:

,

.

For , I employ the substitution , so that and . Also, the interval of integration changes from to , so that

Next, I employ the trigonometric identity :

,

where I have changed the dummy variable from back to .

Therefore, becomes

.

Once again, the fact that the integrand is over an interval of length allows me to shift the interval of integration.

I’ll continue this different method of evaluating this integral in tomorrow’s post.

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by and then employing the substitution (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

So far, I have shown that

,

where , , and is a certain angle (that will soon become irrelevant).

I now write as a new sum by dividing the region of integration:

,

.

For , I employ the substitution , so that and . Also, the interval of integration changes from to , so that

Next, I employ the trigonometric identity :

,

where I have changed the dummy variable from back to .

Therefore, becomes

.

Next, I employ the substitution , so that and the interval of integration changes from to :

.

Almost by magic, the mysterious angle has completely disappeared, making the integral that much easier to compute.

I’ll continue this different method of evaluating this integral in tomorrow’s post.

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by and then employing the substitution (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

So far, I have shown that

.

To simplify the denominator even further, I will combine the two trigonometric terms in the denominator; this is possible because the argument of both the sine and cosine functions are the same. To this end, notice that

,

where

Next, let be the unique angle so that

,

.

With this substitution, we find that

Therefore, the integral may be rewritten as

,

where and .

I’ll continue this different method of evaluating this integral in tomorrow’s post.

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by and then employing the substitution (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

So far, I have shown that

We now employ the substitution , so that . Also, the limits of integration change from to , so that

Next, I’ll divide write by dividing the interval of integration (not to be confused with the and used in the previous method), where

For , I employ the substitution , so that and . Under this substitution, the interval of integration changes from $2\pi \le \theta \le 4\pi$ to $0 \le u \le 2\pi$, and so

Next, I use the periodic property for both sine and cosine — and — to rewrite as

Except for the dummy variable , instead of , we see that is identical to . Therefore,

.

I’ll continue this different method of evaluating this integral in tomorrow’s post.

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by and then employing the substitution (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

For this next technique, I begin by using the trigonometric identities

,

,

.

Using these identities, we obtain

In this way, the exponents have been removed from the denominator, thus making the integrand somewhat less complicated.

I’ll continue this different method of evaluating this integral in tomorrow’s post.

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

In this series, I’ll explore different ways of evaluating this integral.So far in this series, I’ve shown that

This last integral can be evaluated using a standard trick. Let , so that . We differentiate this last equation with respect to :

Employing a Pythagorean identity, we have

Since , we may rewrite this as

Integrating both sides with respect to , we obtain the antiderivative

We now employ this antiderivative to evaluate :

And so, at long last, we have arrived at the solution for the integral . Surprisingly, the answer is independent of the parameter .

These last few posts illustrated the technique that I used to compute this integral for my wife in support of her recent paper in Physical Review A. However, I had more than a few false starts along the way… or, at the time, I thought they were false starts. It turns out that there are multiple ways of evaluating this integral, and I’ll explore another method of attack beginning with tomorrow’s post.